Maximum compression of a spring

  • Thread starter supersingh
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  • #1

Homework Statement




A 50-kg mass is dropped from a 20-m height onto a very long, initially uncompressed spring of force constant k = 100 N/m

Homework Equations


Ueo=Ugf+Uef
F=k(Lenght Final-length initial)


The Attempt at a Solution


I tried using Ueo=Ugf+Uef, and i couldnt figure out what to use for some of the values.
so far i got 1/2*4.9^2=Ugf+Uef, i dont know what to put for the final heights.
 

Answers and Replies

  • #2
malty
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Homework Statement




A 50-kg mass is dropped from a 20-m height onto a very long, initially uncompressed spring of force constant k = 100 N/m

Em but what exactly is the question asking to solve here?
 
  • #3
oh my bad, its asking what is the maximum compression of the spring
 
  • #4
malty
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oh my bad, its asking what is the maximum compression of the spring
Ahh my bad, should have heeded the title a little better.:blushing:

Well, what will determine how much the spring decreases in length by? Using this can you work out the length the spring decreases by?
 
  • #5
i found out that when the object is in equllibrium, the spring goes displaces 4.9 m using
mg=k(l-L) where k is 100 and the mass is 50
 
  • #6
Doc Al
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i found out that when the object is in equllibrium, the spring goes displaces 4.9 m using
mg=k(l-L) where k is 100 and the mass is 50
The equilibrium point is not needed for this problem.

Hint: In calculating the change in gravitational PE, be sure to use the total change in height.
 
  • #7
malty
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i found out that when the object is in equllibrium, the spring goes displaces 4.9 m using
mg=k(l-L) where k is 100 and the mass is 50
Yes but are you sure it's [mg=k(l-L)]

Edit: Ahh I've been beaten me here.
 
  • #8
im pretty sure it is, becuase the problem before it needed that height to calculate the velocity at equillibrium, and i got that
 
  • #9
malty
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im pretty sure it is, becuase the problem before it needed that height to calculate the velocity at equillibrium, and i got that
Maybe, but in this case the equilibrium position doesn't matter, because the most the spring can be compressed is the force of mass that fell onto it.
 
  • #10
Doc Al
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Another hint: This is a conservation of energy problem.
 
  • #11
in the intial situations, there is only intial elastic potential
then for the final, there is final elastic and final gravitational. what value should i use for the height for the gravitational value?
 
  • #12
Doc Al
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Initially, there's only gravitational PE. Hint: If you measure gravitational PE using the lowest position of the mass as the zero point, then in the final position (of maximum compression) there will only be elastic PE.

Another hint: Call the amount by which the spring compresses X.
 
  • #13
wait, i got it, thanks everyone, i now used Ugo=Uef+Ugf
 

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